acids and bases

ACIDS AND BASES
Label the following properties as “A” for acid, “B” for base or “AB” for acids and bases
Tastes bitter
Increases [H+]
Corrosive
Tastes sour
Increase [OH-]
Electrolyte
Neutralizes HCl
Proton acceptor
Proton donor
Neutralizes NaOH
Contains H+ ions
Contains OH- ions
Feels slippery
Turns litmus red
Turns litmus blue
SVANTE ARRHENIUS
NEUTRALIZATION
acids and bases:
EQUATION:
BRONSTED-LOWRY acids and bases:
Match these substances with their chemical formulas, then write as “A” for acid, “B” for base where applicable
____1. Table salt
____2. Water
____3. Stomach acid
____4. Sugar
____5. Limewater
____6. Household ammonia
____7. Hydrogen peroxide
____8. Chalk
____9. Drain cleaner (lye)
____10. Vinegar
____11. Battery acid
____12. Citric acid
a) H2O2
b) CaCO3
c) HC2H3O2
d) H3C6H5O7
e) H2O
f) NH3
g) C12H22O11
i) Ca(OH)2
j) H2SO4
k) NaOH
l) NaCl
m) HCl
ELECTROLYTES
NAMING ACIDS
“ate” becomes “_____ic acid”
“ite” becomes “_____ous acid”
“ide” becomes “hydro_____ic acid
Ion name
chloride
sulfide
cyanide
iodide
nitrate
chlorate
phosphate
sulphate
sulfite
chlorite
nitrite
perchlorate
hypochlorite
permanganate
periodate
phosphite
carbonate
bromate
thiosulfate
iodate
bicarbonate*
hydroxide*
oxide*
peroxide*
* = be careful
Ion
formula
in sulfur compounds, add “ur”
in phosphorus compounds, add “or”
Acid
Formula
Acid
Name
A C I D S
A N D
B A S E S
C A L C U L A T I O N S
What volume of 0.15M sodium hydroxide solution is required to neutralize 275mL of a 0.40 M hydrochloric acid solution? What is the concentration of 20.0 ml of sulphuric acid neutralized by 65 ml of 0.25 M lithium hydroxide solution? ACIDS AND BASES CALCULATIONS (NEUTRALIZATION QUESTIONS) 1) If 35 ml of 0.75M HCl (aq) are neutralized by 85 ml of NaOH (aq), determine
the concentration of the base.
2) What volume of 0.620M phosphoric acid will neutralize 75 ml of 0.250 M
KOH (aq)?
3) What volume of 0.150 M sulphuric acid will neutralize 45 ml of 0.0800M
aluminum hydroxide solution?
One more PRECIPITATION QUESTION… 4) Determine the ion concentrations following the precipitation resulting from the
combination of 150. ml of 0.45 M aluminum sulphate solution and
200. ml of 0.60 M strontium hydroxide solution.
ACIDS
AND
BASES
KEY
Label the following properties as “A” for acid, “B” for base or “AB” for acids and bases
Tastes bitter
Increases [H+]
Corrosive
B
A
AB
A
Tastes sour
B
Increase [OH-]
AB
Electrolyte
B
Neutralizes HCl
B
Proton acceptor
A
Proton donor
A
Neutralizes NaOH
A
Contains H+ ions
B
Contains OH- ions
B
Feels slippery
A
Turns litmus red
B
Turns litmus blue
SVANTE ARRHENIUS
acids and
• acids produce H+ ions in aqueous solutions
• bases produce OH- ions in aqueous solutions
NEUTRALIZATION EQUATION:
ACID + BASE  SALT + WATER
HA + BOH
 BA + HOH
BRONSTED-LOWRY
• acids are proton donors
• bases are proton acceptors
HNO3 (aq) + H2O  NO3-(aq) + H3O+(aq)
NH3 (aq) + H2O  NH4+(aq) + OH-(aq)
acids
and
bases:
bases:
Match these substances with their chemical formulas, then write as “A” for acid, “B” for base
where applicable
L 1. Table salt
a) H2O2
b) CaCO3
E 2. Water
c) HC2H3O2
M 3. Stomach acid A
d) H3C6H5O7
G 4. Sugar
e) H2O
I 5. Limewater B
f) NH3
g) C12H22O11
F 6. Household ammonia B
i) Ca(OH)2
A 7. Hydrogen peroxide A
j) H2SO4
B 8. Chalk
k) NaOH
K 9. Drain cleaner (lye) B
l) NaCl
C 10. Vinegar A
m) HCl
J 11. Battery acid A
D 12. Citric acid A
ELECTROLYTES
An electrolyte is any substance that will conduct electricity when dissolved in water, due to the presence
of free ions. EXAMPLES:
+
−
Dissociation of salts:
NaCl(s) → Na (aq) + Cl (aq)
Strong acids and bases are electrolytes:
HCl (aq) + H2O  Cl-(aq) + H3O+(aq)
NaOH (aq)  Na+(aq) + OH-(aq)
More on this in Chem 12
NAMING
“ate” becomes “_____ic acid”
“ite” becomes “_____ous acid”
“ide” becomes “hydro_____ic acid
Ion
Ion
name
formula
chloride
Clsulfide
S2cyanide
CNiodide
Initrate
NO3chlorate
ClO3phosphate
PO43sulphate
SO42sulfite
SO32chlorite
ClO2nitrite
NO2perchlorate
ClO4hypochlorite
ClOpermanganate
MnO4periodate
IO4phosphite
PO33carbonate
CO32bromate
BrO3thiosulphate
S2O32iodate
IO3bicarbonate*
HCO3hydroxide*
OHoxide*
O2peroxide*
O22* = be careful
ACIDS
KEY
in sulfur compounds, add “ur”
in phosphorus compounds, add “or”
Acid
Acid Name
Formula
HCl
hydrochloric acid
H2S
hydrosulphuric acid
HCN
hydrocyanic acid
HI
hydroiodic acid
HNO3
nitric acid
HClO3
chloric acid
H3PO4
phosphoric acid
H2SO4
sulphuric acid
H2SO3
sulphurous acid
HClO2
chlorous acid
HNO2
nitrous acid
HClO4
perchloric acid
HClO
hypochlorous acid
HMnO4
permanganic acid
HIO4periodic acid
H3PO3
phosphorous acid
H2CO3
carbonic acid
HBrO3
bromic acid
H2S2O3
thiosulphuric acid
HIO3iodic acid
H2CO3
carbonic acid
HOH
water
H2O
water
H2O2
hydrogen peroxide
ACIDS AND BASES CALCULATIONS (NEUTRALIZATION QUESTIONS)
KEY 1) If 35 ml of 0.75M HCl (aq) are neutralized by 85 ml of NaOH (aq), determine
the concentration of the base.
HCl (aq)
(0.75
+ NaOH(aq)
 H2O(l) + NaCl(aq)
)(0.035L) =
2.6X10-2 moles
 2.6X10-2 moles
In any neutralization, Moles H+(aq) = moles OH-(aq)
since H+(aq) + OH-(aq) H2O(l)
Think of this water formation equation as the net equation for a
neutralization reaction.
The ultimate goal is the H +(aq) from the acid and the OH-(aq) from the
base come together to form water, neutralizing the acid and base.
The other product is not represented in the net equation since
Na+(aq) + Cl-(aq) NaCl(aq) should really be written as
Na+(aq) + Cl-(aq) Na+(aq) + Cl-(aq)
since the ions in the salt (ionic compound) are completely dissociated in
aqueous solution.
If moles of acid = moles of base = 2.6X10-2 moles
Then we could complete this question by:
[NaOH] =
= 0.31 M NaOH
Another way to consider completing this question is:
MaVa = MbVb
(0.75
)(0.035L) = Mb (0.085L)
Mb = 0.31 M NaOH
NEUTRALIZATION QUESTIONS
KEY
page two
2) What volume of 0.620M phosphoric acid will neutralize 75 ml of 0.250 M
KOH (aq)?
BE CAREFUL HERE!
First consider the balanced neutralization reaction:
H3PO4(aq)

+ 3KOH(aq)
(0.075L)(0.250
0.00625 moles
3H2O(l) + K3PO4 (aq)
)=
0.01875 moles
divide by 3
Then we can determine the volume of H3PO4(aq) :
=
X 0.00625moles = 0.01008 L = 10.08 mL
Due to SF we must report 10. mL
NOTE!
Another way to consider doing this question is recognizing that when
H3PO4 and KOH react, the H+ and OH- are in a 3:1 ratio and therefore
before we begin this method:
MaVa = MbVb
We must use 3 times the molarity for H3PO4 (0.620M X 3), in order to
have it equal the RS of the equation, MbVb:
(1.86
)Va = (0.250M) (0.075L)
Va= 10.08 mL
NEUTRALIZATION QUESTIONS
KEY
page three
3) What volume of 0.150 M sulphuric acid will neutralize 45 ml of 0.0800M
aluminum hydroxide solution?
3H2SO4(aq)
+ 2Al(OH)3(aq)
 6H2O(l) + Al2(SO4)3 (aq)
Consider looking at the question in the following way.
Consider the dissociation of the acid and the base in aqueous solution:
H2SO4(aq) 
2H+(aq) + SO4-2(aq)
0.150M
0.300M
and
Al(OH)3(aq)  Al+3(aq) + 3OH-(aq)
0.0800M
0.240M
Then we can use
MaVa = MbVb
Keeping in mind that MaVa = moles H+(aq) and Mb Vb = moles OH-(aq)
SO:
(0.300M) Va = (0.240 M)(45 ml)
Va = 36 mL
Note the variety of ways in which we can arrive at the correct
solutions for these types of problems!!!
You need to develop this flexibility in finding these solutions in order to
survive the 8 to 10 week acid and base unit in Chemistry 12!!!!!!!!!
One more PRECIPITATION QUESTION…
KEY
page four
4) Determine the ion concentrations following the precipitation resulting from the
combination of 150. ml of 0.45 M aluminum sulphate solution and
200. ml of 0.60 M strontium hydroxide solution.
DILUTION
Al(SO4)3(aq)
Sr(OH)2(aq)
= 0.19M
= 0.34M
Note that I am using the TOTAL final volume of 350 mL in
my M1V1 = M2V2 dilution calculation. M2 =
DISSOCIATION
Al2(SO4)3(aq)  2Al
0.19M
+3
0.38M
(aq)
+ 3SO4-2(aq)
0.57M
Sr(OH)2(aq)  Sr+2aq) + 2OH-(aq)
0.34M
0.34M
0.68M
PRECIPITATION
Possible precipitates:
Al(OH)3
and SrSO4
BOTH ARE POSSIBLE PRECIPITATES!
Al+3(aq)
0.38M
0.23M
+
3OH-(aq)  Al(OH)3(s)
Sr+2aq)
1.14M
0.68M
0.34M
0.57M
+ SO4-2aq)
0.34M
0.57M
 SrSO4(s)
[Al+3] = 0.38 (avail) – 0.23 M (reacted) = 0.15 M (excess)
[OH-] = 0 M (limiting reagent; totally used up in precipitate formation)
[Sr+2]= 0 M (limiting reagent; totally used up in precipitate formation)
[SO4-2] = 0.57 (avail) – 0.34 M (reacted) = 0.123 M (excess)

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